Gujarat BoardEnglish MediumSTD 11 ScienceMATHSBINOMIAL THEOREM1 Mark
Question
Find the coefficient of $a^4$ in the product $(1 + 2a)^4 (2 – a)^5$ using the binomial theorem.
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Answer
We first expand each of the factors of the given product using Binomial Theorem. Here, we have
$\Longrightarrow(1+2 a)^4={ }^4 C_0+{ }^4 C_1(2 a)+{ }^4 C_2(2 a)^2+{ }^4 C_3(2 a)^3+{ }^4 C_4(2 a)^4$
$=1+4(2 a)+6\left(4 a^2\right)+4\left(8 a^3\right)+16 a^4$
$=1+8 a+24 a^2+32 a^3+16 a^4$
and also $(2-\mathrm{a})^5={ }^3 \mathrm{C}_0(2)^5-{ }^5 \mathrm{C}_1(2)^4(a)+{ }^5 \mathrm{C}_2(2)^3(a)^2-{ }^5 \mathrm{C}_3(2)^2(a)^3+{ }^5 \mathrm{C}_4(2)(a)^4-{ }^5 \mathrm{C}_5(a)^5$
$=32-80 a+80 a^2-40 a^3+10 a^4-a^5$
Therefore, $(1+2 a)^4(2-a)^5=\left(1+8 a+24 a^2+32 a^3+16 a^4\right)\left(32-80 a+80 a^2-40 a^3+10 a^4-a^5\right)$
The complete multiplication of the two brackets need not be carried out.
We write only those terms which involve $a^4$
This can be done if we note that $a^r \cdot a^{4-r}=a^4$. The terms containing $a^4$ are
$1\left(10 a^4\right)+(8 a)\left(-40 a^3\right)+\left(24 a^2\right)\left(80 a^2\right)+\left(32 a^3\right)(-80 a)+\left(16 a^4\right)(32)=-438 a^4$
Therefore, the required coefficient of $\mathrm{a}^4$ in the given product is -438
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